This problem set accompanies Chapter 2 (Solar Energy) of the textbook
Energy and the New Reality, Volume 2:
Carbon-Free Energy Supply
by
L. D. Danny Harvey
Department of Geography
University of Toronto
harvey@geog.utoronto.ca
Published by Earthscan (London, UK)
Homepage: www.earthscan.co.uk/?tabid=101807
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Problem Set 1 – Solar Energy
Assigned:
Due:
Taken up:
PART 1, Computation of incident solar irradiance
Question 1
In this question you will learn how to compute the intensity of solar radiation (the irradiance)
incident on a flat surface with any orientation at any latitude at any time of day on any day of the
year, for clear sky conditions. To do this, you will use the set of equations that is given and
explained in Box 2.1 of the draft textbook.
The first part of this exercise requires you to carefully type a series of equations into the Excel
spreadsheet that will be emailed to you – quite a few equations, in fact, but they are all given to
you or arise as very simple modifications to the equations that you are given. You will have to
convert back and forth between angles in degrees and radians, so you should figure out the factor
that you need to use in each case first and test your result with your hand calculator before
proceeding. You will also have to go from the cosine of an angle to the angle itself using the
arccos function, which is simply acos in Excel. The angle that acos gives you will be in radians
(which is something that you can verify yourself).
It’s also useful to keep a few definitions in mind:
1. zenith angle means the angle measured from the vertical (so solar zenith angle means the
angle between the position of the sun in the sky and the vertical)
2. azimuth simply means the horizontal direction (where 0° is due south, -90° is due east,
+90° is due west , and ±180° is due north)
3. the angle of incidence of solar radiation on a surface is the angle between the radiation
and the normal (the perpendicular) to the surface
4. solar declination is the latitude that is directly under sun at solar noon (it varies
seasonally from about 23.3°S to 23.3°N)
5. albedo refers to the fraction of solar radiation incident on a surface that is reflected
You will do your calculations for a horizontal module, for a tilted solar module facing due south,
and for a sun-tracking module (always oriented perpendicular to the sun’s rays). The steps in the
computation are as follows:
1.
2.
In cells H12 and H13, compute the value of pi and the factor that you will need to convert
back and forth between angles in radians and in degrees (see the Worksheet on how to
compute pi).
In cell H18, compute the angle of the Earth in its orbit from the vernal equinox, based on
the number of days from the vernal equinox that is given in cell H9. To do this, assume
that the Earth moves at a constant angular speed in its orbit (this is not exactly correct),
and treat a year as being exactly 365.25 days in length. In cells H19 and H20, compute
the relative Earth-sun distance and the solar declination (in radians), respectively, using
the equations given in Box 2.1. The equation in Box 2.1 gives the declination in radians,
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3.
and this is how you will use it in the subsequent equations, but in cell H21 show what
your computed declination is in degrees, so that you can check that your answer is
reasonable. In cell H24, set the module inclination (for the inclined- or tilted-module
case) equal to the latitude minus the declination. If the module inclination is set equal to
the latitude minus declination at the beginning of each day, what would the angle of
incidence of the sun’s radiation on the module be at solar? (draw a diagram to show the
derivation of your answer). (Solar noon is when the sun is at its highest point and directly
over the longitude in question (that is, due south if you’re looking at it from the northern
hemisphere))
Column A starting in row 41 gives time of day in terms of hours from solar noon (so -4 is
4 hours before solar noon, +4 is four hours after solar noon). In column B, convert the
hours from solar noon to an angle (in radians) from solar noon (You’ll have to think
through the steps required to do this, and then type in the required equation. Think of
simple cases to test your result, and try again if your answer fails the test). Indicate how
you arrived at the equation to go from hours from solar noon to the angle from solar
noon. For this and other equations, don’t forget the ‘$’ that needs to go in front of the row
number for cells that you want to hold fixed when you drag the equation down after
typing it into the top row.
4.
In column D, compute cos (where  is the solar zenith angle in radians). Type: =Max(0,
Eq 2.4 of Box 2.1). Recall that Excel assumes that the angles being given to it for
trigonometric functions are in radians. By taking the maximum of zero and the result of
Eq 2.4, the result will be zero when the sun is below the horizon. This has to be the case
since, in the equation for solar irradiance, the irradiance on a horizontal plane varies with
cos, and we certainly want the irradiance to be zero when the sun is below the horizon!
5.
In column E, compute the solar zenith angle in degrees. You will not actually use this
column in any of your calculations, but it will be helpful in visualizing (and checking)
your results if you can see what the zenith angle is in more intuitive units, and you will
use it in making graphs.
6.
Columns F, G, and H already have in them the equations for the cosine of the solar
azimuth, for solar azimuth (in radians), and for cos for the case of a module inclined at
an angle (given in cell H24) equal to the latitude minus the solar declination and with an
azimuth of zero ( is the angle between the sun and a line normal (perpendicular) to the
module). cos is based on Eq 2.6 of Box 2.1 rather than Eq 2.8, so that the equation is
still valid when there is a non-zero module azimuth. Correct values will appear once all
the steps up to this stage have been correctly implemented. The following slight
adjustments to the equations were made to get the correct answer for all possible
circumstances, or at least to have clean results for plotting later:
(i)
(ii)
The cosine of the solar azimuth (column F) is computed as Min(1.0, Eq 2.7 of
Box 2.1) because, with roundoff error when Excel does the calculations, it is
possible that Excel gives a value for cos a ever so slightly greater than 1.0 when
the sun is due south. If that happens, one cannot take the arcos of cos a, which is
needed to get a itself.
We use the arccos function, acos, to get the solar azimuth in radians from the
computed cosine of the solar azimuth (Column G), but for certain times we have
to use –acos instead of acos, so that the azimuth angle has the correct sign (the
arccos function in Excel always gives a positive angle).
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(iii)
(iv)
For cos (column H) we use Max(0.0, Eq 2.6 of Box 2.1). A negative cos
would correspond to the sun being below the “horizon” formed by the tilted
module, in the same way that a negative cos would correspond to the sun being
below the Earth’s horizon, and since the module (like the Earth) is not
transparent, we have to set cos = 0 when the sun would be shining on the back
side of the module so that the irradiance computed later with Eq 2.10 is zero.
The equation for cos* is multiplied by cos/(cos + 0.0000000001). When cos
is not zero, this term will be indistinguishable from 1.0 (so it changes nothing),
and when cos=0 it will be equal to zero. In this way, we force cos* to be zero
whenever the sun is below the horizon (that is, whenever cos is zero), which in
turn guarantees that IDR (Eq 2.10 of Box 2.1) will be zero whenever the sun is
below the horizon (otherwise – the following could happen: if the module is
tilted to the east or west and the sun is just below the Earth’s horizon (cos set to
0), cos could still be > 0, depending on the tilt of the module and how much
below the horizon the sun is. Since the computed irradiance on tilted modules
depends directly on cos*, we could end up with a computed non-zero irradiance
at times when the sun is below the horizon if we do not force cos* to be zero at
such times).
Note: All of the above manoeuvres are what computer programmers would refer to as
“tricks”, but there is nothing deceptive about them!
7.
Columns J, L, M, N and O already contain the equations for optical air mass m (Eq 2.18
of Box 2.1), DR (directly transmitted fraction of solar radiation), SF (forward scattered
fraction), SB (back scattered fraction for radiation reflected up from the surface), and
AW (fractional absorption of solar radiation by water vapour). m, DR, SF and AW depend on
the solar zenith, while SB applies to diffuse radiation, which comes from all directions
and so does not depend on the solar zenith angle, but rather, is based on the average
zenith angle.
8.
9.
In column P, compute the surface albedo
In column Q, compute the common factor
 Qs
 2
d

(1  O3)(1  Aw )

that appears in the equations for IDR, IDf and IR (Eqs. (2.10) to (2.12). O3 is the fraction of
solar radiation absorbed by ozone in the stratosphere, which you can assume to be 0.03 at
all times.
10.
Compute D1 and D2 in columns R and S, and the sum in column T. Note that we use both
s (the value computed in column P) and s-bar (from cell H26) in the equation for D2.
11.
Horizontal module case: Derive and show the simplified versions of Eqs. (2.10)-(2.12)
that arise for horizontal modules, and use the simplified equations in your spreadsheet in
columns V, W, and X.
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12.
Inclined module case: Use Eqs. (2.10)-(2.12), except that a slight problem arises for IDf
when cos=0, since we have to divide by it. This might not seem to matter, since the only
time it happens, the sun is below the horizon (i.e., there is no sunlight anyway). However,
it causes problems if we want to compute daily averages or graph the results. We would
like energy terms to be zero when the sun is below the horizon. Also, it would be nice to
type only one equation that works for all situations. We can make the computations
automatically do this without noticeably changing the results when the sun is above the
horizon by computing cos*/cos as cos*/( cos + 0.0000000001). This works because
your cos* should be zero whenever your cos is zero, and when it isn’t zero, the
0.0000000001 has a negligible effect on the answer. When cos is zero, the
0.0000000001 prevents us from dividing by zero.
13.
Sun-tracking module case. When the module tracks the sun, the module is always
perpendicular to the sun’s rays. This allows you to simplify Eqs. (2.10) and (2.12) a little,
but you have to deduce what the appropriate module inclination is at any given time.
Drawing a picture should help you figure this out. You’ll have to use the trick that you
used in step 12 for IDF to force the answer for IDR to be zero when the sun is below the
horizon. You’ll also need to make sure that you never divide by zero in the equation for
IDF, but this will require a different trick than used for IDR. Show the simplified equations,
including the tricks that you have to embed inside the equation for IDR and IDF (if you
can’t think of a trick for IDF, just manually set the value to zero where required).
Having done all this, do or answer the following:
(i)
(ii)
(iii)
(iv)
(v)
Prepare a graph showing how DR, SF, and AW vary with the solar zenith angle. You only
need to plot the data on one side of solar noon, since the data are (or should be!)
symmetric about solar noon. Comment briefly on the variation of SF and AW vs that of DR,
with solar zenith angle, and propose a simple explanation for the relationships [3 marks]
How does your computed surface albedo vary with solar zenith angle?
Prepare a graph showing the diurnal variation of total solar radiation (that is, the sum of
direct, diffuse, and reflected radiation) on a horizontal module, on the inclined module
with the inclination set equal to the latitude minus the declination, and on the suntracking module. Compute the daily average solar radiation for each case, and place this
information on your graph sheet but below the graph area. Comment on what you see [6
marks].
Copy your worksheet and set the module inclination to 90 (vertical), then copy the
altered worksheet 7 more times, so that you have a total of 8 copies with a vertical
module. For the first four, choose module azimuths corresponding to due south, east,
west and north. These are the summer cases. For the other four, choose the same four
module azimuths, but set the number of days since the vernal equinox to 270. This will
give you four winter cases. Your results for the “Inclined Module Case” in the 8 altered
worksheets now correspond to the solar radiation incident on walls or windows facing
each of the four cardinal directions, for winter and summer.
Prepare a graph for one of the directions comparing the diurnal variation total incident
radiation in summer and winter. Once you’re satisfied with the look of the graph, copy it
three times and simply change the worksheets that each copy is using for the data that are
plotted (by clicking on each line that is plotted on your graph and changing the worksheet
name referred to in the window at the top), so as to quickly generate graphs for the other
three directions with minimal effort (and don’t forget to modify the title of each copied
graph). This will be easiest if, when you originally made your eight copies, the name that
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you put in the tab at the bottom of each worksheet is short, such as WS, WE, WW, WN
for the four winter cases. Hand in the four graphs.
For each case, briefly explain the reason for the difference between summer and winter (7
marks; you can discuss east and west orientations together). For a rectangular building,
what would be the best orientation so as to minimize summer over-heating? What would
be the best orientation so at to maximize solar heat gain in winter?
Note 1: “orientation” of a building means how the long-axis is oriented – for example
running north-south or in some other direction. Note 2: By comparison with Figure 2.4 of
the textbook, you should be able to deduce what your graphs should look like. If your
graphs are obviously wrong, then base your discussion of the differences seen between
winter and summer on Fig 2.4, not on your results.
Hand in the top half of your first worksheet (i.e., the portion that you see in the template attached
to this problem set), along with your graphs and the answers to the questions.
PART 2, Analysis of the Economics and Embodied Energy of Solar Energy
Question 2
In this question you will calculate the cost of providing photovoltaic power for various
assumptions concerning component costs and efficiencies. Use the cost and efficiency data
provided in the worksheet “PV Cost” to work out the cost of PV electricity, following the steps
indicated in the Excel spreadsheet. You will need to carefully read and study Box 2.2 of the
textbook. Pay close attention to your units throughout – doing so will go a long way toward
helping you to see what formulae to use. Include all of the capital costs that you see
given in the spreadsheet. Those that are given as $/m2 can be converted to $/kWp DC in the same
way that module costs (as $/m2) are converted to $/kWp DC in Box 2.2, then plugged into Eq.
(B2.2.3). Any costs that are already given as $/kW DC can be directly plugged into Eq. (2.26).
For the sake of transparency, I am asking you to separately compute the capital cost and O&M
contributions to the cost of electricity. The O&M is a fixed fraction (each year) of the total capital
cost, where total capital cost includes the ID (indirect cost) term. Ignore insurance.
What effect does doubling the efficiency of the module have on the cost of electricity, and why?
Compare this effect with the effect of cutting the module cost in half, and comment.
If the mean annual solar irradiance on a horizontal module is 200 W/m2, what would be the
annual capacity factor of the PV system? What would the irradiance be in units of kWh/m2/yr?
Hand in your worksheet and your response to the above questions.
Question 3
In this question you will work out the cost of a hybrid system consisting of PV modules and
natural gas turbines as the backup, similarly to what is asked in the wind energy problem set. As
in the wind problem set, you will compute the cost of electricity from solar alone and from
natural gas alone. Then, you will compute the average cost of electricity in the hybrid system in
two ways – one based on a simple weighting of the separate solar and natural gas electricity costs,
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and the other based on the total system capital costs and the total annual generation of electricity.
We will optionally allow for a small reduction in the efficiency of the natural gas system (from
0.55 to 0.50) when it serves as backup for fluctuating solar power compared to when it alone is
supplying all of the electricity. You can assume that the average demand satisfied by the hybrid
system or natural gas-only system is 65% of the peak demand, and that no solar electricity is ever
wasted (i.e., that it can all be used).
The module and BOS efficiencies have not been included in the worksheet for this question. This
is because, with the information given to you, you don’t need to know the efficiencies. Figure out
and write the equation that you will need to use to determine the cost of electricity, and explain
why the module and BOS efficiencies do not appear in the equation. The easiest way to do this is
to probably think about the cost equation for a fossil fuel system.
I’m also asking you to compute the C tax required to equalize the cost of electricity from the
solar-natural gas hybrid and from natural gas alone. This is equivalent to the cost of avoided C
emission, which you computed for wind energy in Problem Set 1.
After you have completed the calculations (in all the yellow cells in the worksheet), answer the
following questions:
1.
Explain why the module and BOS efficiencies do not appear in the cost equation for solar
electricity this time.
What kind of natural gas system are we implicitly assuming if we give it an efficiency (for
generating electricity) of 0.55?
Explain why the cost of the hybrid system as computed with method two is greater than the
cost as computed from a simple weighting of the separate solar and natural gas costs.
For the case where we’ve allowed for ~10% reduction in the original efficiency of the
natural gas system (from 0.55 to 0.50), you’ll find that the reduction in CO2 emissions
decreases by more than 10% (compared to the case where PV is added with no reduction in
efficiency). Explain why this is so.
What happens to the required C tax as the cost of natural gas increases, assuming no
reduction in the efficiency of the natural gas system? What effect does the small reduction
in the efficiency of the natural gas system have on the required C tax, especially at high
natural gas costs? Explain your results. How would characterize this situation?
Comment on the required C tax when natural gas costs $15/GJ (it has fluctuated between
$4/GJ to $12/GJ recently) and the installed cost of the PV system drops in half (to
$2000/kWp-AC). Given that total annual per capita C emissions in developed countries are
about 2-5 tC/yr/person, are the required C taxes large or small?
2.
3.
4.
5.
6.
Question 4
The production of PV modules and the related electronic systems requires a lot of energy – the
so- called embodied energy. In the worksheet for this question, you are given embodied energies
in terms of primary energy. Compute the following:
1.
2.
The embodied primary energy averaged over all the kWhs produced by a solar
system during its lifetime.
The amount of primary energy that is saved per kWh of electricity that is generated,
assuming that the generated solar electricity displaces natural gas-generated
electricity with the fixed efficiency and transmission loss given in the worksheet.
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3.
The number of years that it takes for the PV system to generate enough electricity,
and thereby save enough primary energy at the fossil fuel powerplant, to pay back the
amount of primary energy required to make the solar system.
This time I’m leaving it up to the student to figure out the sequence of calculations that will be
needed to answer these questions, but I’ll give a couple of hints: You’ll first have to figure out
what the embodied energy is per peak kW of AC capacity, and you’ll have to figure out how
many kWh of AC electricity will be generated by the PV system over its lifetime per peak kW of
AC capacity.
Question 5
The final question considers the merits of covering half of the available roof area with dedicated
PV modules and half of the available area with solar thermal collectors, versus covering the entire
available roof area with various hybrid PV/thermal modules. Included in the work sheet are the
electrical and thermal efficiencies for stand-alone and hybrid systems. Calculate the savings in
primary energy, given the indicated electrical powerplant and heating system efficiencies, and
assuming that addition of solar thermal or solar PV has no effect on the heating system or
powerplant efficiencies, respectively. Comment on the merits of having dedicated PV and solar
thermal systems (each covering half of the available roof area) vs a hybrid system covering all of
the available roof area.
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